# Tensor product: numerical example

Reference: Moore, Thomas A., A General Relativity Workbook, University Science Books (2013) – Chapter 6; Problem 6.4.

It’s useful to do a tensor calculation using actual numbers, just to see how they work. Suppose we have a vector with components

$\displaystyle A^{i}=\left[1,2,-1,0\right] \ \ \ \ \ (1)$

where the units are metres, and a second vector given by

$\displaystyle B^{i}=\left[3,-1,0,-2\right] \ \ \ \ \ (2)$

where the units are ${\mbox{seconds}^{-1}}$.

Since the units are different (even using relativistic units where length = time, since vector ${B^{i}}$ has units of 1/time), we cannot add these vectors. However, we can multiply them to get a rank-2 tensor:

$\displaystyle M^{ij}=A^{i}B^{j}=\left[\begin{array}{cccc} 3 & -1 & 0 & -2\\ 6 & -2 & 0 & -4\\ -3 & 1 & 0 & 2\\ 0 & 0 & 0 & 0 \end{array}\right] \ \ \ \ \ (3)$

The first row consists of ${A^{0}}$ multiplied by each element of ${B}$ in turn, and so on (the indices ${i}$ and ${j}$ each run for 0 to 3). The units of each element of ${M^{ij}}$ are ${\mbox{m}\cdot\mbox{s}^{-1}}$.

The question in Moore asks if this product is commutative, which isn’t really the right thing to ask, since obviously ${A^{i}B^{j}=B^{j}A^{i}}$. However, the product is not symmetric, in the sense that ${A^{i}B^{j}\ne A^{j}B^{i}}$. That is, it matters which vector is used to determine the row index and which the column index.

The trace of ${M}$ is given by

 $\displaystyle M_{\;\; i}^{i}$ $\displaystyle =$ $\displaystyle g_{ij}M^{ij}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle g_{ij}A^{i}B^{j} \ \ \ \ \ (5)$

This is just the scalar product ${\mathbf{A}\cdot\mathbf{B}}$. If we use the special relativity metric ${\eta_{ij}}$, then

$\displaystyle \eta_{ij}A^{i}B^{j}=-\left(1\times3\right)+\left(-1\times2\right)+\left(-1\times0\right)+\left(-2\times0\right)=-5 \ \ \ \ \ (6)$

We could also have obtained this result from the matrix ${M^{ij}}$ directly by calculating ${\eta_{00}M^{00}+\eta_{11}M^{11}=-5}$ (the other two diagonal elements are zero).